Since HCl and NaOH react in a 1:1 ratio, 0.200 L x 2.2 M NaOH = 0.440 mols will react with 0.200 x 3.9 = 1.17 mols HCl to form (ONLY) 0.44 mol NaCl.
M = mols/L soln (careful--the solution volume will be 300 + 200 = 500 mL or 0.5L).
to 200 mL of 2.2 molar NaOH solution, what
will be the molarity of NaCl in the resulting
solution?
Answer in units of M
M = mols/L soln (careful--the solution volume will be 300 + 200 = 500 mL or 0.5L).
HCl + NaOH -> NaCl + H2O
First, let's calculate the number of moles of HCl and NaOH:
1. Moles of HCl = (volume in liters) x (molarity)
Moles of HCl = (300 mL / 1000 mL/L) x 3.9 M
Moles of HCl = 0.3 L x 3.9 M
Moles of HCl = 1.17 moles
2. Moles of NaOH = (volume in liters) x (molarity)
Moles of NaOH = (200 mL / 1000 mL/L) x 2.2 M
Moles of NaOH = 0.2 L x 2.2 M
Moles of NaOH = 0.44 moles
Since the balanced equation shows a 1:1 stoichiometric ratio between HCl and NaCl, the moles of NaCl formed will be the same as the moles of NaOH added.
3. Moles of NaCl = Moles of NaOH
Moles of NaCl = 0.44 moles
Finally, we need to calculate the molarity of NaCl in the resulting solution:
4. Molarity of NaCl = (moles of NaCl) / (volume in liters)
Molarity of NaCl = 0.44 moles / (300 mL + 200 mL) / 1000 mL/L
Molarity of NaCl = 0.44 moles / 0.5 L
Molarity of NaCl = 0.88 M
Therefore, the molarity of NaCl in the resulting solution is 0.88 M.
Step 1: Determine the moles of HCl and NaOH used
Moles of HCl = Volume of HCl solution (in liters) * Molarity of HCl
Moles of NaOH = Volume of NaOH solution (in liters) * Molarity of NaOH
First, we need to convert the volumes from milliliters to liters:
Volume of HCl solution = 300 mL = 300 / 1000 = 0.3 L
Volume of NaOH solution = 200 mL = 200 / 1000 = 0.2 L
Next, we calculate the moles of HCl and NaOH:
Moles of HCl = 0.3 L * 3.9 M = 1.17 moles
Moles of NaOH = 0.2 L * 2.2 M = 0.44 moles
Step 2: Determine the reaction between HCl and NaOH
HCl + NaOH -> NaCl + H2O
From the balanced equation, we see that 1 mole of HCl reacts with 1 mole of NaOH to produce 1 mole of NaCl.
Step 3: Determine the moles of NaCl formed
Since the reaction between HCl and NaOH is a 1:1 ratio, the number of moles of NaCl formed is equal to the number of moles of NaOH used.
Moles of NaCl formed = Moles of NaOH used = 0.44 moles
Step 4: Determine the molarity of NaCl in the resulting solution
Molarity of NaCl = Moles of NaCl formed / Volume of resulting solution (in liters)
The resulting solution is obtained by adding the volumes of the HCl and NaOH solutions together:
Volume of resulting solution = Volume of HCl solution + Volume of NaOH solution
Volume of resulting solution = 0.3 L + 0.2 L = 0.5 L
Now we can calculate the molarity of NaCl:
Molarity of NaCl = 0.44 moles / 0.5 L = 0.88 M
Therefore, the molarity of NaCl in the resulting solution is 0.88 M.